[EM] De-Cloned Kendall-tau Example

[Forest Simmons]

I appreciate very much Kevin Venzke's recent tour de force towards
classification of methods by their criteria compliances ... finding seven
3-candidate, 4-faction profiles that distinguish all of the well known
methods by their basic compliances.

I see this as an opportunity to put the de-cloned Kendall-tau metric
through its paces.

First a simple example with factions suitable for hand computation:

4ABC3+3BCA+2CAB

We need favorite and anti-favorite counts for the de-cloning:

The respective favorite counts f, are 4, 3, and 2. The respective a
anti-favorite counts f', are 3, 2, and 4.

The respective costs of the basic decloned order swaps are ...
AB to BA 4×2 from f(A)×f'(B)
BA to AB 3×3 from f(B)×f'(A)
BC to CB 3×4 from f(B)×f'(C)
CB to BC 2×2 from f(C)×f'(B)
CA to AC 2×3 from f(C)× f'(A)
AC to CA 4×4 from f(A)×f'(C)

The respective costs from faction ...
ABC to BCA 8+16
BCA to ABC 6+9
[Round trip 39]
BCA to CAB 12+9
CAB to BCA 8+4
[Round trip 33]
CAB to ABC 6+4
ABC to CAB 12+16
[Round trip 38]

Our method is to remove pairs of ballots that are as far apart as possible
until only one faction remains:

ABC and BCA are the furthest apart ballots with a round trip distance of
39.

Removing 3  ballots from each of these factions leaves ...

ABC+2CAB

After removing one ballot from each of these two remaining factions, we see
that the winning faction is CAB.

So this is one way to use the de-cloned Kendall-tau distance.

To do Kevin's seven examples I will need some triple A batteries for my
calculator.

The most interesting part for me is that the geometrically derived
preferences would be ...

The 4 members of the ABC faction should geometrically prefer the CAB
faction over the BCA faction because they are closer to it ... distance 38
verses 39.

The 3 members of the BCA faction should prefer the CAB faction over the ABC
faction ... distance 33 versus 39.

The two member CAB faction should prefer BCA to ABC .... distance 33 vs 39.

So geometric preferences are

2 CAB>BCA>ABC
3 BCA>CAB>ABC
4 ABC>CAB>BCA

The CAB faction is the Condorcet faction.

In other words sincere preferences seem to be

4 A>C>B
3 B>C>A
2 C>B>A

It looks like insincere or mistaken burials of C by A and A by C.

Any other thoughts?
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